Divide the following complex numbers. $ \dfrac{27-5i}{5-2i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${5+2i}$ $ \dfrac{27-5i}{5-2i} = \dfrac{27-5i}{5-2i} \cdot \dfrac{{5+2i}}{{5+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(27-5i) \cdot (5+2i)} {(5-2i) \cdot (5+2i)} = \dfrac{(27-5i) \cdot (5+2i)} {5^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(27-5i) \cdot (5+2i)} {(5)^2 - (-2i)^2} = $ $ \dfrac{(27-5i) \cdot (5+2i)} {25 + 4} = $ $ \dfrac{(27-5i) \cdot (5+2i)} {29} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({27-5i}) \cdot ({5+2i})} {29} = $ $ \dfrac{{27} \cdot {5} + {-5} \cdot {5 i} + {27} \cdot {2 i} + {-5} \cdot {2 i^2}} {29} $ Evaluate each product of two numbers. $ \dfrac{135 - 25i + 54i - 10 i^2} {29} $ Finally, simplify the fraction. $ \dfrac{135 - 25i + 54i + 10} {29} = \dfrac{145 + 29i} {29} = 5+i $